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two positive point charges of 12uc and 8uc are 10cm apart. the work done in bringing them 4cm closer is?

two positive point charges of 12uc and 8uc are 10cm apart. the work done in bringing them 4cm closer is?

Grade:12th pass

6 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
5 years ago
Hii

Find out the change in the potential energy of the system and take the difference. You will get the answer.


Best of Luck
Adarsh srivastav
11 Points
3 years ago
q1=12×10^-6,q2=8×10^-6.
Wkt,potential when the are at R1(10 cm)
V1=kq1q2/R1=86.4×10^-1~=86×10^-1.
V2=kq1q2/R2=216×10^-1.
W.D=V2-V1=(216×10^-1)-(86×10^-1)=130×10^-1
                                                                =13J
Purba bandyopadhyay
15 Points
3 years ago
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta
askIITians Faculty 82 Points
one year ago
 
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta
askIITians Faculty 82 Points
one year ago
HiiFind out the change in the potential energy of the system and take the difference. You will get the answer.Best of Luck
 
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem below.
 
Initial distance between the charges, d1 = 10 cm = 0.1 m
Final distance between the charges, d2 = 10 cm – 4 cm = 6 cm = 0.06 m
Hence, W = 9 x 109 x 12 x 10-6 x 8 x 10-6 (1/0.06 – 1/0.1)
               = 5.76 J
 
Hope it helps.
Thanks and regards,
Kushagra

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